$f(x)=\dfrac{1}{x^2}$ Find the third degree Taylor polynomial, centered at $x=1$, of $f$. Choose 1 answer: Choose 1 answer: (Choice A) A $1+2(x-1)+\frac{6{{(x-1)}^{2}}}{2!}+\frac{24{{(x-1)}^{3}}}{3!}$ (Choice B) B $1-2(x-1)+\frac{6{{(x-1)}^{2}}}{2!}-\frac{24{{(x-1)}^{3}}}{3!}$ (Choice C) C $1-2(x+1)+\frac{6{{(x+1)}^{2}}}{2!}-\frac{24{{(x+1)}^{3}}}{3!}$ (Choice D) D $ (x-1)-\frac{2(x-1)^2}{2!}+\frac{6{{(x-1)}^{3}}}{3!}-\frac{24{{(x-1)}^{4}}}{4!}$
Explanation: First, find the first three derivatives of $~f(x)=\dfrac{1}{x^2}=x^{-2}\,$. ${f}\,^\prime(x)=-2x^{-3}=-\frac{2}{{{x}^{3}}}$ ${f}\,^{\prime\prime}(x)=6x^{-4}=\dfrac{6}{{{x}^{4}}}$ ${f}\,^{\prime\prime\prime}(x)=-24x^{-5}=-\dfrac{24}{{{x}^{5}}}$ Then let $~x=1~$ in the original function and in these derivatives to get the coefficients for $~{{T}_{3}}\left( x \right)\,$, the third-degree Taylor polynomial for $~f\left( x \right)\,$. $ f(1)=1\,;~~~~{f}\,^\prime(1)=-2\,;~~~~{f}\,^{\prime\prime}(1)=6\,;~~~~{f}\,^{\prime\prime\prime}(1)=-24$ Use these coefficients in the equation for the Taylor polynomial of degree $~3\,$. ${{T}_{3}}\left( x \right)=f(1)+{f}\,^\prime(1)(x-1)+\frac{{f}\,^{\prime\prime}(1){{(x-1)}^{2}}}{2!}+\frac{{f}\,^{\prime\prime\prime}(1){{(x-1)}^{3}}}{3!}$ Hence, ${{T}_{3}}\left( x \right)=1-2(x-1)+\frac{6{{(x-1)}^{2}}}{2!}-\frac{24{{(x-1)}^{3}}}{3!}$